Assume we have an abelian category which has fibre products. Let $f:X\to Z$ and $g:Y\to Z$ be two morphisms and let $(W,p,q)$ be their fibre product with $p:W\to X$, $q:W\to Y$. If the category is a module category over some ring and $f$ is an epimorphism, then we get an induced short exact sequence $0\to K\to W\to Y\to 0$, where $K$ is the kernel of $f$. My question is if this is true in arbitrary abelian categories with fibre products. I am able to show that the kernel of $q$ is isomorphic to the kernel of $f$, so it would suffice to show that $q$ is an epimorphism if $f$ is an epimorphism. But I don't know how to manage that.
If the category is small, then I know that it is equivalent to some module category. So I am searching for a completely arrow-theoretic proof.
Every abelian category has fiber products. In fact, fiber products can always be constructed via equalizers and products, and equalizers in $\mathsf{Ab}$-categories may be reduced to kernels.
Your question is why epimorphisms in abelian categories are stable under pullback. You can find the proof in Mac Lane's CWM, Chapter VIII, Prop. 4.2.
By the way, you can also assumethat your abelian category is small (consider the closure of $X,Y,\dotsc,f,g,\dotsc$ under direct sums, kernels and cokernels) and then use the embedding theorem of Freyd-Mitchell.